Answer :
Final answer:
The change in entropy (ΔS) for the vaporization of 0.50 mol ethanol is calculated by dividing the product of the molar heat of vaporization (39.3 kJ/mol) and the amount of substance (0.50 mol) by the boiling point in Kelvin (351.45 K).
Explanation:
The enthalpy of vaporization is the heat required to vaporize one mole of a substance at its boiling point under standard pressure. For ethanol, this value is given as 39.3 kJ/mol.
To calculate the change in entropy (ΔS) for the vaporization of ethanol, we use the formula ΔS = ΔH/T, where ΔH is the enthalpy change and T is the temperature in Kelvin. Converting the boiling point of ethanol from °C to K, we have 78.3°C + 273.15 = 351.45 K. For 0.50 mol of ethanol, the ΔS would be (0.50 mol × 39.3 kJ/mol) / 351.45 K.